What is the integral of 1/(sqrt(x^2-2)) ?

The integral of 1/(sqrt(x^2-2)) is ln|sqrt(1/2 (x^2-2))+1/(sqrt(2))x|+C

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integral of 1/(sqrt(x^2-2))

\int \frac{1}{x ^{2} - 2} d x

ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x + C

Solution steps

\int \frac{1}{x ^{2} - 2} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{2} x)

= ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))

Simplify

ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)) : ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x

= ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x

Add a constant to the solution

= ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x + C

\int (x + \frac{1}{4 x}) d x

h (t) = t - 2

\int x (x + 6)^{7} d x

\frac{\partial}{\partial y} (cos (x) + x cos (y) - y)

n = 0 \sum \infty (\frac{2}{4})^{n}

What is the integral of 1/(sqrt(x^2-2)) ?
The integral of 1/(sqrt(x^2-2)) is ln|sqrt(1/2 (x^2-2))+1/(sqrt(2))x|+C